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CSE 143 Lecture 16

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Exercise: Dice rolls

Write a method diceRoll that accepts an integer parameter representing a number of 6-sided dice to roll, and output all possible combinations of values that could appear on the dice.

diceRoll(2);diceRoll(3);

[1, 1, 1]

[1, 1, 2]

[1, 1, 3]

[1, 1, 4]

[1, 1, 5]

[1, 1, 6]

[1, 2, 1]

[1, 2, 2]

...

[6, 6, 4]

[6, 6, 5]

[6, 6, 6]

Examining the problem

We want to generate all possible sequences of values.

for (each possible first die value):

for (each possible second die value):

for (each possible third die value):

...

print!

This is called a depth-first search

How can we completely explore such a large search space?

Backtracking

backtracking: Finding solution(s) by trying partial solutions and then abandoning them if they are not suitable.

a "brute force" algorithmic technique (tries all paths)

often implemented recursively

Applications:

producing all permutations of a set of values

parsing languages

games: anagrams, crosswords, word jumbles, 8 queens

combinatorics and logic programming

Backtracking algorithms

A general pseudo-code algorithm for backtracking problems:

Explore(choices):

if there are no more choices to make: stop.

else, for each available choice C:

Make choice C.

Explore the remaining choices.

Un-make choice C, if necessary. (backtrack!)

A decision tree

...

...

...

...

...

...

...

...

Private helpers

Often the method doesn't accept the parameters you want.

So write a private helper that accepts more parameters.

Extra params can represent current state, choices made, etc.

public int methodName(params):

...

return helper(params, moreParams);

private int helper(params, moreParams):

...

(use moreParams to help solve the problem)

Exercise solution

// Prints all possible outcomes of rolling the given

// number of six-sided dice in [#, #, #] format.

public static void diceRolls(int dice) {

List<Integer> chosen = new ArrayList<Integer>();

diceRolls(dice, chosen);

}

// private recursive helper to implement diceRolls logic

private static void diceRolls(int dice,

List<Integer> chosen) {

if (dice == 0) {

System.out.println(chosen); // base case

} else {

for (int i = 1; i <= 6; i++) {

chosen.add(i); // choose

diceRolls(dice - 1, chosen); // explore

chosen.remove(chosen.size() - 1); // un-choose

}

}

}

Exercise: Dice roll sum

Write a method diceSum similar to diceRoll, but it also accepts a desired sum and prints only combinations that add up to exactly that sum.

diceSum(2, 7);diceSum(3, 7);

[1, 6]

[2, 5]

[3, 4]

[4, 3]

[5, 2]

[6, 1]

[1, 1, 5]

[1, 2, 4]

[1, 3, 3]

[1, 4, 2]

[1, 5, 1]

[2, 1, 4]

[2, 2, 3]

[2, 3, 2]

[2, 4, 1]

[3, 1, 3]

[3, 2, 2]

[3, 3, 1]

[4, 1, 2]

[4, 2, 1]

[5, 1, 1]

New decision tree

...

Optimizations

We need not visit every branch of the decision tree.

Some branches are clearly not going to lead to success.

We can preemptively stop, or prune, these branches.

Inefficiencies in our dice sum algorithm:

Sometimes the current sum is already too high.

(Even rolling 1 for all remaining dice would exceed the desired sum.)

Sometimes the current sum is already too low.

(Even rolling 6 for all remaining dice would exceed the desired sum.)

When finished, the code must compute the sum every time.

(1+1+1 = ..., 1+1+2 = ..., 1+1+3 = ..., 1+1+4 = ..., ...)

Exercise solution, improved

public static void diceSum(int dice, int desiredSum) {

List<Integer> chosen = new ArrayList<Integer>();

diceSum2(dice, desiredSum, chosen, 0);

}

private static void diceSum(int dice, int desiredSum,

List<Integer> chosen, int sumSoFar) {

if (dice == 0) {

if (sumSoFar == desiredSum) {

System.out.println(chosen);

}

} else if (sumSoFar <= desiredSum &&

sumSoFar + 6 * dice >= desiredSum) {

for (int i = 1; i <= 6; i++) {

chosen.add(i);

diceSum(dice - 1, desiredSum, chosen, sumSoFar + i);

chosen.remove(chosen.size() - 1);

}

}

}

Backtracking strategies

When solving a backtracking problem, ask these questions:

What are the "choices" in this problem?

What is the "base case"? (How do I know when I'm out of choices?)

How do I "make" a choice?

Do I need to create additional variables to remember my choices?

Do I need to modify the values of existing variables?

How do I explore the rest of the choices?

Do I need to remove the made choice from the list of choices?

Once I'm done exploring, what should I do?

How do I "un-make" a choice?

Exercise: Permutations

Write a method permute that accepts a string as a parameter and outputs all possible rearrangements of the letters in that string. The arrangements may be output in any order.

Example:permute("TEAM")outputs the followingsequence of lines:

Examining the problem

We want to generate all possible sequences of letters.

for (each possible first letter):

for (each possible second letter):

for (each possible third letter):

...

print!

Each permutation is a set of choices or decisions:

Which character do I want to place first?

Which character do I want to place second?

...

solution space: set of all possible sets of decisions to explore

Decision tree

...

Exercise solution

// Outputs all permutations of the given string.

public static void permute(String s) {

permute(s, "");

}

private static void permute(String s, String chosen) {

if (s.length() == 0) {

// base case: no choices left to be made

System.out.println(chosen);

} else {

// recursive case: choose each possible next letter

for (int i = 0; i < s.length(); i++) {

char c = s.charAt(i); // choose

s = s.substring(0, i) + s.substring(i + 1);

chosen += c;

permute(s, chosen); // explore

s = s.substring(0, i) + c + s.substring(i + 1);

chosen = chosen.substring(0, chosen.length() - 1);

} // un-choose

}

}

Exercise solution 2

// Outputs all permutations of the given string.

public static void permute(String s) {

permute(s, "");

}

private static void permute(String s, String chosen) {

if (s.length() == 0) {

// base case: no choices left to be made

System.out.println(chosen);

} else {

// recursive case: choose each possible next letter

for (int i = 0; i < s.length(); i++) {

String ch = s.substring(i, i + 1); // choose

String rest = s.substring(0, i) + // remove

s.substring(i + 1);

permute(rest, chosen + ch); // explore

}

} // (don't need to "un-choose" because

} // we used temp variables)

Exercise: Combinations

Write a method combinations that accepts a string s and an integer k as parameters and outputs all possible k -letter words that can be formed from unique letters in that string. The arrangements may be output in any order.